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A proof of the irrationality of \sqrt{2} just to try \LaTeX in WordPress

Assume \sqrt{2} is rational.

Then \exists a and b such that \sqrt{2} = \frac{a}{b}
where a and b are integers and \frac{a}{b} is in lowest terms.

\Rightarrow 2b^2 = a^2

\Rightarrow a^2 is even

\Rightarrow a is even

\Rightarrow a is divisible by 2

\Rightarrow \exists k such that a = 2k

substitute 2k for a in the assumption to get
\sqrt{2} = \frac{2k}{b}

\Rightarrow 2 = \frac{(2k)^2}{b^2} = \frac{4k^2}{b^2}

\Rightarrow 2k^2 = b^2

\Rightarrow b is even, similarly.

\Rightarrow \frac{a}{b} is not in lowest terms. \blacksquare

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Posted in Mathematics
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